开场：正定矩阵的收尾

这一讲先用十分钟补完正定矩阵，再进入主题。补的部分回答三个自然的问题。

正定矩阵的逆还是正定吗

是。理由用特征值最快：若 $A$ 的特征值是 $\lambda_1,\dots,\lambda_n>0$，则 $A^{-1}$ 的特征值是 $1/\lambda_1,\dots,1/\lambda_n$，仍然全为正。逆矩阵对称，因此也正定。

两个正定矩阵之和还是正定吗

是。不需要知道 $A+B$ 的特征值或主元，直接看二次型：

$$x^{\mathsf T}(A+B)x=x^{\mathsf T}Ax+x^{\mathsf T}Bx.$$

右边是两个正数之和，对任何 $x\neq 0$ 都为正。

这条证明也说明了一件事：遇到正定性问题，先试二次型，它往往绕开了特征值和主元。

$A^{\mathsf T}A$ 什么时候正定

设 $A$ 是 $m\times n$ 的长方矩阵。它本身通常不是方阵，谈不上正定；但 $A^{\mathsf T}A$ 是 $n\times n$ 对称矩阵，正是最小二乘法里的核心组合。

$$x^{\mathsf T}A^{\mathsf T}Ax=(Ax)^{\mathsf T}(Ax)=\lVert Ax\rVert^2\ge 0.$$

所以 $A^{\mathsf T}A$ 永远半正定。它正定的充要条件是 $\lVert Ax\rVert^2>0$ 对所有 $x\neq 0$ 成立，也就是 $Ax\neq 0$，即 $A$ 的零空间只有零向量。

用秩来表述：

$$A^{\mathsf T}A\text{ 正定}\iff A\text{ 的列线性无关}\iff \operatorname{rank}A=n.$$

这也解释了为什么最小二乘的正规方程在列无关时可以放心求解：$A^{\mathsf T}A$ 不仅可逆，而且正定。

补充：正定矩阵在数值计算中还有一个重要性质——消元时不需要换行。主元不会变成零或极小值，因此 $A=LU$ 的分解稳定。这使正定矩阵成为数值线性代数里最「好算」的一类矩阵。

时间轴

相似矩阵的定义

设 $A,B$ 都是 $n\times n$ 方阵。若存在可逆矩阵 $M$，使得

$$B=M^{-1}AM,$$

则称 $A$ 与 $B$ 相似。

这个定义看起来像一次「换基」：同一个线性变换，在不同的基下写成不同的矩阵。$M$ 就是从旧基到新基的过渡矩阵。

最重要的相似：对角化

若 $A$ 有 $n$ 个线性无关的特征向量，把它们作为列组成 $S$，则

$$S^{-1}AS=\Lambda.$$

所以 $A$ 相似于对角矩阵 $\Lambda$。在所有与 $A$ 相似的矩阵里，$\Lambda$ 是形式最简单的一个。

但 $S$ 不是唯一的选择。任取一个可逆的 $M$，$M^{-1}AM$ 都会得到另一个与 $A$ 相似的矩阵，只是通常不再是对角阵。

相似矩阵共享特征值

这是本讲的主定理：

相似矩阵有相同的特征值。

证明

设 $Ax=\lambda x$，且 $B=M^{-1}AM$。在 $A$ 与 $x$ 之间插入 $MM^{-1}$：

$$A(MM^{-1})x=\lambda x.$$

左边重新结合为 $(AM)(M^{-1}x)$，再在两边左乘 $M^{-1}$：

$$M^{-1}AM\,(M^{-1}x)=\lambda\,(M^{-1}x).$$

即

$$B(M^{-1}x)=\lambda(M^{-1}x).$$

所以 $\lambda$ 也是 $B$ 的特征值，对应的特征向量是 $M^{-1}x$。

特征值不变，特征向量被 $M^{-1}$ 搬到了新基下。 这也符合直觉：换基不改变线性变换本身，只改变它的坐标表示。

一个可手算的例子

取

$$A=\begin{bmatrix}2&1\\1&2\end{bmatrix},\qquad M=\begin{bmatrix}1&1\\0&1\end{bmatrix},\qquad M^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}.$$

先算 $AM$：

$$AM=\begin{bmatrix}2&1\\1&2\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix} =\begin{bmatrix}2&3\\1&3\end{bmatrix}.$$

再左乘 $M^{-1}$：

$$B=M^{-1}AM= \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{bmatrix}2&3\\1&3\end{bmatrix} =\begin{bmatrix}1&0\\1&3\end{bmatrix}.$$

$A$ 的特征值是 $3$ 与 $1$。$B$ 的迹是 $4$，行列式是 $3$，特征值同样是 $3$ 与 $1$。

$B$ 看起来和 $A$ 很不一样，但它们属于同一个相似类。

相似类的图景

固定特征值 $3$ 与 $1$，所有具有这两个特征值的 $2\times2$ 矩阵构成一个家族，家族成员两两相似。家族的「代表」是对角矩阵

$$\begin{bmatrix}3&0\\0&1\end{bmatrix}.$$

除此之外还有上三角、下三角、各种非对称形式——只要特征值仍是 $3$ 与 $1$，就都能通过某个 $M$ 与 $A$ 相连。

坏情形：特征值重复

若 $\lambda_1=\lambda_2$，事情变得微妙：矩阵可能没有足够多的特征向量，因而不可对角化。

一个孤独的矩阵

考虑 $4I=\begin{bmatrix}4&0\\0&4\end{bmatrix}$。对任何可逆 $M$，

$$M^{-1}(4I)M=4M^{-1}M=4I.$$

它只与自己相似。在所有特征值为 $4,4$ 的矩阵中，$4I$ 自成一类。

另一个家族

矩阵

$$\begin{bmatrix}4&1\\0&4\end{bmatrix}$$

的特征值也是 $4,4$，但它只有一个线性无关的特征向量：零空间由 $\begin{bmatrix}1\\0\end{bmatrix}$ 张成，维数为 $1$。它不可对角化，因此不可能与 $4I$ 相似。

所有其余特征值为 $4,4$ 的矩阵都落在同一个相似类里，而 $\begin{bmatrix}4&1\\0&4\end{bmatrix}$ 是这个类里最接近对角的形式。

更多同族成员

只要保持迹为 $8$、行列式为 $16$，就得到同族的矩阵，例如

$$\begin{bmatrix}5&1\\-1&3\end{bmatrix},\qquad \begin{bmatrix}4&17\\0&4\end{bmatrix}.$$

它们都只有一个特征向量，彼此相似。

数特征向量还不够

一个自然的猜想是：特征值相同、特征向量个数相同，就应该相似。这个猜想是错的。

考虑两个 $4\times4$ 矩阵，特征值全是 $0$。

第一个：两个独立的 $2\times2$ 块

$$J_1=\begin{bmatrix} 0&1&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0 \end{bmatrix}.$$

它的秩为 $2$，零空间维数为 $4-2=2$，有两个特征向量。

第二个：一个 $3\times3$ 块加一个 $1\times1$ 块

$$J_2=\begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}.$$

它的秩同样是 $2$，零空间维数同样是 $2$，也有两个特征向量。

但 $J_1$ 与 $J_2$ 不相似。区别不在数量，而在块的大小：$J_1$ 是 $2+2$，$J_2$ 是 $3+1$。这说明要完整刻画相似类，必须知道特征向量的「归属结构」，而不只是总数。

Jordan 块与 Jordan 定理

Jordan 块

一个 Jordan 块 $J_i$ 形如

$$J_i=\begin{bmatrix} \lambda_i&1&&\\ &\lambda_i&\ddots&\\ &&\ddots&1\\ &&&\lambda_i \end{bmatrix},$$

对角线上是重复的特征值，超对角线上全是 $1$，其余为零。

关键性质：一个 Jordan 块只有一个特征向量。块越大，缺失的特征向量越多。

Jordan 定理

每个方阵 $A$ 都相似于某个 Jordan 矩阵 $J$，$J$ 由若干 Jordan 块沿对角线拼成。

$$J=\begin{bmatrix} J_1&&\\ &\ddots&\\ &&J_d \end{bmatrix}.$$

两个要点：

1. 块的数量等于线性无关特征向量的个数，因为每块贡献一个。
2. 块的大小记录了特征向量的缺失方式，这正是相似类之间的区别。

若所有特征值互不相同，每个块都是 $1\times1$，$J$ 就是对角矩阵，$d=n$。所以对角化是 Jordan 形的特例。

为什么现在不强调 Jordan 形

Jordan 形在代数上很漂亮：它给出了相似类的完整分类。但它对数值极其敏感——只要特征值有微小扰动，重根就会分开，块结构随之改变，秩也会跳变。因此实际计算中很少真的去求 Jordan 形。

课堂的判断是：Jordan 形是理解相似类的正确语言，但不再是这门课的高潮；真正的主角是下一讲的奇异值分解。

本讲结论

• 正定矩阵的逆、和仍然是正定；$A^{\mathsf T}A$ 恒为半正定，列无关时为正定。
• 相似的定义是 $B=M^{-1}AM$，等价于同一个线性变换换一组基。
• 相似矩阵共享全部特征值，特征向量按 $M^{-1}$ 变换。
• 特征值重复时，相似类会分裂；$4I$ 自成一类，其余矩阵构成另一类。
• 仅统计特征向量个数不足以判断相似，还要看 Jordan 块的结构。
• 每个方阵都相似于一个 Jordan 矩阵；特征值互异时它退化为对角阵。

理解检验

1. 为什么 $A^{\mathsf T}A$ 总是半正定？它什么时候正定？

参考答案：对任意 $x$，

$$x^{\mathsf T}A^{\mathsf T}Ax=(Ax)^{\mathsf T}(Ax)=\lVert Ax\rVert^2\ge0,$$

所以二次型从不为负，$A^{\mathsf T}A$ 半正定（它本身对称：$(A^{\mathsf T}A)^{\mathsf T}=A^{\mathsf T}A$）。要正定，需要 $\lVert Ax\rVert^2>0$ 对所有 $x\ne0$ 成立，即 $Ax\ne0$ 对所有非零 $x$ 成立，也就是 $N(A)=\{\mathbf{0}\}$。这等价于 $A$ 的列线性无关，即 $\operatorname{rank}A=n$。列相关的方向会让 $Ax=0$，二次型在该方向取零，此时只是半正定。

2. 证明相似矩阵有相同的特征值，并写出特征向量之间的对应关系。

参考答案：设 $B=M^{-1}AM$，且 $Ax=\lambda x$，$x\ne0$。在 $A$ 与 $x$ 之间插入 $MM^{-1}$：

$$A(MM^{-1})x=\lambda x \;\Longrightarrow\; (AM)(M^{-1}x)=\lambda x .$$

两边左乘 $M^{-1}$：

$$M^{-1}AM\,(M^{-1}x)=\lambda\,(M^{-1}x),$$

即 $B(M^{-1}x)=\lambda(M^{-1}x)$。因为 $M^{-1}$ 可逆且 $x\ne0$，所以 $M^{-1}x\ne0$，它是 $B$ 的属于同一特征值 $\lambda$ 的特征向量。因此特征值集合相同，特征向量按 $x\mapsto M^{-1}x$ 搬到新基下。

3. 为什么 $4I$ 只与自己相似？

参考答案：对任何可逆 $M$，

$$M^{-1}(4I)M=4M^{-1}M=4I .$$

所以与 $4I$ 相似的所有矩阵都只能是 $4I$ 本身，它的相似类只有一个成员。直观上，$4I$ 在每个方向上都把向量拉伸 $4$ 倍，换任何基都不改变这一表示；它的特征空间是整个平面，没有「特殊方向」可被换基改变。

4. 两个 $4\times4$ 矩阵都有两个特征向量，为什么仍可能不相似？

参考答案：特征向量个数相同只是零空间维数相同，不足以刻画 Jordan 块的结构。正文的两个矩阵

$$J_1=\begin{bmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{bmatrix}, \qquad J_2=\begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$$

秩都为 $2$，零空间维数都为 $2$，各有两列特征向量；但 $J_1$ 是两块 $2\times2$，$J_2$ 是一块 $3\times3$ 加一块 $1\times1$。区分方法是看幂的秩：$J_1^2=0$，秩为 $0$；而 $J_2^2$ 在第 $(1,3)$ 位仍为 $1$，秩为 $1$。相似变换保持 $\operatorname{rank}((A-\lambda I)^k)$，所以两者不可能相似。

5. 特征值互不相同时，Jordan 形是什么样子？为什么？

参考答案：此时 Jordan 形就是对角矩阵，每个块都是 $1\times1$，共有 $n$ 个块。原因是：不同特征值对应的特征向量线性无关，$n$ 个互不相同的特征值给出 $n$ 个线性无关的特征向量，矩阵可对角化；而 Jordan 块的大小由特征向量的缺失决定，没有缺失就没有大于 $1$ 的块。于是 $J=\Lambda$，对角化只是 Jordan 形的特例。

与前后课程的关系

上一讲用二次型与海森矩阵刻画了正定矩阵；这一讲先补完它的运算性质，再进入相似与 Jordan 形。下一讲转向奇异值分解：它不需要矩阵对称，也不需要可对角化，却对所有矩阵都成立。

Finishing positive definiteness

The lecture spends its first ten minutes closing three loose ends.

Is the inverse of a positive definite matrix positive definite?

Yes. Eigenvalues give the shortest proof: if $A$ has eigenvalues $\lambda_1,\dots,\lambda_n>0$, then $A^{-1}$ has eigenvalues $1/\lambda_1,\dots,1/\lambda_n$, still positive. The inverse is symmetric, so it is positive definite.

Is the sum of two positive definite matrices positive definite?

Yes. There is no need to know the eigenvalues or pivots of $A+B$; use the quadratic form:

$$x^{\mathsf T}(A+B)x=x^{\mathsf T}Ax+x^{\mathsf T}Bx.$$

The right side is a sum of two positive numbers for every $x\neq 0$.

The proof also carries a lesson: when a positive-definiteness question appears, try the quadratic form first. It often bypasses eigenvalues and pivots entirely.

When is $A^{\mathsf T}A$ positive definite?

Let $A$ be $m\times n$. It is usually rectangular, so it is not itself positive definite. But $A^{\mathsf T}A$ is $n\times n$ and symmetric — the central object of least squares.

$$x^{\mathsf T}A^{\mathsf T}Ax=(Ax)^{\mathsf T}(Ax)=\lVert Ax\rVert^2\ge 0.$$

So $A^{\mathsf T}A$ is always positive semidefinite. It is positive definite exactly when $\lVert Ax\rVert^2>0$ for every $x\neq 0$, that is, when $Ax\neq 0$, meaning the null space of $A$ is trivial.

In rank terms:

$$A^{\mathsf T}A\text{ positive definite}\iff A\text{ has independent columns}\iff \operatorname{rank}A=n.$$

This is why the normal equations are safe to solve when the columns are independent: $A^{\mathsf T}A$ is not merely invertible but positive definite.

Background: positive definite matrices also need no row exchanges during elimination. Pivots never become zero or dangerously small, so $A=LU$ is numerically stable. It is the best-behaved class in numerical linear algebra.

Timeline

The definition of similarity

Let $A$ and $B$ be $n\times n$. They are similar if there is an invertible $M$ with

$$B=M^{-1}AM.$$

The definition is a change of basis: the same linear transformation written in a different basis, where $M$ is the change-of-basis matrix.

The most important similarity: diagonalization

If $A$ has $n$ independent eigenvectors, put them in the columns of $S$. Then

$$S^{-1}AS=\Lambda.$$

So $A$ is similar to the diagonal matrix $\Lambda$, the simplest member of its family.

But $S$ is not the only choice. Any invertible $M$ produces another matrix $M^{-1}AM$ similar to $A$, usually no longer diagonal.

Similar matrices share eigenvalues

The main theorem:

Similar matrices have the same eigenvalues.

Proof

Suppose $Ax=\lambda x$ and $B=M^{-1}AM$. Insert $MM^{-1}$ between $A$ and $x$:

$$A(MM^{-1})x=\lambda x.$$

Regroup the left side as $(AM)(M^{-1}x)$ and multiply both sides on the left by $M^{-1}$:

$$M^{-1}AM\,(M^{-1}x)=\lambda\,(M^{-1}x).$$

That is,

$$B(M^{-1}x)=\lambda(M^{-1}x).$$

So $\lambda$ is also an eigenvalue of $B$, with eigenvector $M^{-1}x$.

The eigenvalues stay fixed; the eigenvectors are moved by $M^{-1}$. This matches intuition: changing basis does not change the transformation, only its coordinate representation.

A hand-checkable example

Take

$$A=\begin{bmatrix}2&1\\1&2\end{bmatrix},\qquad M=\begin{bmatrix}1&1\\0&1\end{bmatrix},\qquad M^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}.$$

First compute $AM$:

$$AM=\begin{bmatrix}2&1\\1&2\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix} =\begin{bmatrix}2&3\\1&3\end{bmatrix}.$$

Then multiply on the left by $M^{-1}$:

$$B=M^{-1}AM= \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{bmatrix}2&3\\1&3\end{bmatrix} =\begin{bmatrix}1&0\\1&3\end{bmatrix}.$$

The eigenvalues of $A$ are $3$ and $1$. The trace of $B$ is $4$ and its determinant is $3$, so its eigenvalues are again $3$ and $1$.

$B$ looks nothing like $A$, yet they belong to the same similarity class.

The picture of a similarity class

Fix the eigenvalues $3$ and $1$. Every $2\times2$ matrix with these eigenvalues forms one family, and any two members are similar. The representative is the diagonal matrix

$$\begin{bmatrix}3&0\\0&1\end{bmatrix}.$$

Upper triangular, lower triangular, and asymmetric forms all join the family as long as the spectrum stays $\{3,1\}$.

The bad case: repeated eigenvalues

When $\lambda_1=\lambda_2$, something subtler happens: the matrix may not have enough eigenvectors to diagonalize.

A matrix alone in its class

Consider $4I=\begin{bmatrix}4&0\\0&4\end{bmatrix}$. For any invertible $M$,

$$M^{-1}(4I)M=4M^{-1}M=4I.$$

It is similar only to itself. Among all matrices with eigenvalues $4,4$, it forms a class of one.

The other family

The matrix

$$\begin{bmatrix}4&1\\0&4\end{bmatrix}$$

also has eigenvalues $4,4$, but only one independent eigenvector: the null space of $A-4I$ is spanned by $\begin{bmatrix}1\\0\end{bmatrix}$, dimension one. It is not diagonalizable, so it cannot be similar to $4I$.

Every other matrix with eigenvalues $4,4$ lies in one big class, and $\begin{bmatrix}4&1\\0&4\end{bmatrix}$ is its closest-to-diagonal representative.

More members

Keep the trace at $8$ and the determinant at $16$, and you stay in the same class, for example

$$\begin{bmatrix}5&1\\-1&3\end{bmatrix},\qquad \begin{bmatrix}4&17\\0&4\end{bmatrix}.$$

Both have a single eigenvector and are similar to each other.

Counting eigenvectors is not enough

A natural guess is that equal eigenvalues plus equal eigenvector counts imply similarity. The guess is false.

Take two $4\times4$ matrices whose eigenvalues are all zero.

First: two independent $2\times2$ blocks,

$$J_1=\begin{bmatrix} 0&1&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0 \end{bmatrix}.$$

Its rank is $2$, so its nullity is $4-2=2$: two eigenvectors.

Second: one $3\times3$ block and one $1\times1$ block,

$$J_2=\begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}.$$

Its rank is again $2$, its nullity again $2$, and it also has two eigenvectors.

Yet $J_1$ and $J_2$ are not similar. The difference is not the count but the block structure: $J_1$ is $2+2$, $J_2$ is $3+1$. A complete description of a similarity class needs to know how the eigenvectors are grouped, not just how many there are.

Jordan blocks and Jordan's theorem

The Jordan block

A Jordan block $J_i$ has the form

$$J_i=\begin{bmatrix} \lambda_i&1&&\\ &\lambda_i&\ddots&\\ &&\ddots&1\\ &&&\lambda_i \end{bmatrix},$$

with a repeated eigenvalue on the diagonal, ones on the superdiagonal, and zeros elsewhere.

The key property: a Jordan block has exactly one eigenvector. The larger the block, the more eigenvectors are missing.

The theorem

Every square matrix $A$ is similar to a Jordan matrix $J$, built from Jordan blocks along the diagonal.

$$J=\begin{bmatrix} J_1&&\\ &\ddots&\\ &&J_d \end{bmatrix}.$$

Two consequences:

1. The number of blocks equals the number of independent eigenvectors, since each block contributes one.
2. The block sizes record how eigenvectors are missing, and that is exactly what distinguishes similarity classes.

If all eigenvalues are distinct, every block is $1\times1$, so $J$ is diagonal and $d=n$. Diagonalization is the special case.

Why Jordan form is no longer the climax

Jordan form is algebraically beautiful: it classifies similarity completely. But it is numerically fragile. A tiny perturbation splits repeated eigenvalues, changes the rank, and alters the block structure.

The lecture's verdict: Jordan form is the right language for understanding similarity classes, but it is no longer the centerpiece of the course. The next lecture's singular value decomposition is.

Takeaways

• The inverse and the sum of positive definite matrices are positive definite; $A^{\mathsf T}A$ is always semidefinite and definite exactly when $A$ has independent columns.
• Similarity means $B=M^{-1}AM$, a change of basis for one linear transformation.
• Similar matrices share all eigenvalues; eigenvectors transform by $M^{-1}$.
• Repeated eigenvalues split the spectrum into several similarity classes; $4I$ is a class of one.
• Counting eigenvectors does not determine similarity; Jordan block structure does.
• Every square matrix is similar to a Jordan matrix, which reduces to a diagonal matrix when eigenvalues are distinct.

Check your understanding

1. Why is $A^{\mathsf T}A$ always positive semidefinite, and when is it positive definite?

Reference answer: For every $x$,

$$x^{\mathsf T}A^{\mathsf T}Ax=(Ax)^{\mathsf T}(Ax)=\lVert Ax\rVert^2\ge0,$$

so the quadratic form is never negative and $A^{\mathsf T}A$ is positive semidefinite (it is also symmetric, since $(A^{\mathsf T}A)^{\mathsf T}=A^{\mathsf T}A$). It is positive definite exactly when $\lVert Ax\rVert^2>0$ for all $x\ne0$, i.e. $Ax\ne0$ for every nonzero $x$, which says $N(A)=\{\mathbf{0}\}$. Equivalently, the columns of $A$ are linearly independent, $\operatorname{rank}A=n$. A dependent direction makes $Ax=0$ and the quadratic form vanish there, leaving only semidefiniteness.

2. Prove that similar matrices have the same eigenvalues and state how their eigenvectors correspond.

Reference answer: Let $B=M^{-1}AM$ and $Ax=\lambda x$ with $x\ne0$. Insert $MM^{-1}$ between $A$ and $x$:

$$A(MM^{-1})x=\lambda x \;\Longrightarrow\; (AM)(M^{-1}x)=\lambda x .$$

Left-multiplying by $M^{-1}$,

$$M^{-1}AM\,(M^{-1}x)=\lambda\,(M^{-1}x),$$

i.e. $B(M^{-1}x)=\lambda(M^{-1}x)$. Since $M^{-1}$ is invertible and $x\ne0$, we have $M^{-1}x\ne0$, so it is an eigenvector of $B$ for the same $\lambda$. The eigenvalue set is unchanged and eigenvectors move by $x\mapsto M^{-1}x$ in the new basis.

3. Why is $4I$ similar only to itself?

Reference answer: For any invertible $M$,

$$M^{-1}(4I)M=4M^{-1}M=4I .$$

So every matrix similar to $4I$ equals $4I$ itself; its similarity class has a single member. Intuitively, $4I$ stretches every direction by the same factor $4$, and no change of basis alters that representation — its eigenspace is the whole space, so there is no distinguished direction for a change of basis to move.

4. Two $4\times4$ matrices both have two eigenvectors. Why can they still fail to be similar?

Reference answer: Equal eigenvector counts only mean equal nullity; they do not capture the Jordan block structure. The two matrices from the text,

$$J_1=\begin{bmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{bmatrix}, \qquad J_2=\begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0\end{bmatrix},$$

both have rank $2$ and nullity $2$, hence two eigenvectors each. But $J_1$ is two $2\times2$ blocks while $J_2$ is one $3\times3$ plus one $1\times1$. The ranks of powers distinguish them: $J_1^2=0$ has rank $0$, whereas $J_2^2$ has a $1$ in position $(1,3)$ and rank $1$. Similarity preserves $\operatorname{rank}((A-\lambda I)^k)$, so the two cannot be similar.

5. What does the Jordan form look like when all eigenvalues are distinct, and why?

Reference answer: It is diagonal, with every block $1\times1$ — that is, $n$ blocks. Distinct eigenvalues have linearly independent eigenvectors, so $n$ distinct eigenvalues supply $n$ independent eigenvectors and the matrix is diagonalizable. Jordan block sizes measure how eigenvectors are missing; with no deficiency, no block is larger than $1\times1$. Thus $J=\Lambda$, and diagonalization is the special case of Jordan form.

How this connects to the course

The previous lecture characterized positive definiteness through quadratic forms and Hessians; this one finishes its algebraic properties, then introduces similarity and Jordan form. The next lecture turns to the singular value decomposition, which requires neither symmetry nor diagonalizability and applies to every matrix.